I have asked this questions previously but I did not understand the responses. I only have the knowledge of AP calculus. Fermat's spiral has the equation $$r^2=a^2 \theta$$ and I am trying to find the $a$ value in the equation so when $r=10$, the arc will be $27.47$ long.
I am projecting the equation onto a cone with a height of $10$ and radius of $10$ and I want the average angle of inclination of the spiral to be $20$ degrees on the cone. That is where the $27.47$ came from. Using the length of a polar curve equation, I came up with the equation $$ 27.47=∫_0^{10^2/a^2} \sqrt{a^2 θ+ \frac{a^2}{4θ}}\, d\theta $$ . The upper bound is from solving for \theta when $r=10$. I do not know how to integrate this. If you can please help it would be greatly appreciated.